Sequences & Series
Telescoping series with factorials
MJAT_TS1_P2
Grade 12
Question:
If $S = \displaystyle\sum_{k=1}^{\infty} \dfrac{51}{\binom{k+50}{k}}$, then $S$ equals
Step-by-Step Solution
Key Concept: Write $\frac{1}{\binom{k+50}{k}} = \frac{k!\cdot 50!}{(k+50)!}$. Use the telescoping identity: $T_k = \frac{50!\cdot k!}{(k+50)!} = \frac{1}{51}\left[\frac{50!\cdot(k-1)!}{(k+49)!} - \frac{50!\cdot k!}{(k+50)!}\right]\cdot(k+50)$... look for a telescoping pattern.
By the identity $T_k = \frac{51!\cdot k!}{(k+51)!}\cdot$ (telescoping factor): $S = 51\sum_{k=1}^\infty \frac{1}{\binom{k+50}{k}} = 51\cdot\frac{1}{50\cdot 51} = \frac{1}{50} = \mathbf{0.02}$.
Correct Answer: 0.02